为什么在堆中创建对象或堆栈中的临时对象时，隐式构造函数将结构中的 POD 初始化为零?

The standard and the C++ book say that the default constructor for class type members is called by the implicit generated default constructor, but built-in types are not initialized. However, in this test program I get unexpected results when allocating an object in the heap or when using a temporary object:

#include<iostream>


struct Container
{
    int n;
};

int main()
{
    Container c;
    std::cout << "[STACK] Num: " << c.n << std::endl;

    Container *pc = new Container();
    std::cout << "[HEAP]  Num: " << pc->n << std::endl;
    delete pc;

    Container tc = Container();
    std::cout << "[TEMP]  Num: " << tc.n << std::endl;

}

我得到这个输出:

[STACK] Num: -1079504552
[HEAP]  Num: 0
[TEMP]  Num: 0

这是编译器特有的行为吗?我真的不打算依赖这个，但我很想知道为什么会发生这种情况，特别是对于第三种情况.

Type x;

Type x = {}; // only works for struct/class without constructor
Type x = Type();
Type x{}; // C++11 only

struct X { int x; X(); };

X::X() {}

X::X() : x() {}

new int;

new int();

Type x();

Type x = Type();

Type x{};